WORTH 22 POINTS please help me :)

Mathematics

1 answer:

GarryVolchara [31]3 years ago

8 0

ANSWER

on a it would be 24 because 34 is not right if you add 12+12. B would be 60$ is the cost of five tickets

Step-by-step explanation:

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Answers choices <1 <4 <2 <3

Maslowich

Answer:

B) 4

Step-by-step explanation:

1. It is either 3 or 4, since those are only two angles comparing the lighthouse and the boat.

2. The angle of depression is noted below the horizontal and above the actual line, and out of 3 and 4, 4 is the only angle that is below its corresponding horizontal.

So, the angle of depression from the lighthouse to the boat is 4.

7 0

3 years ago

What's the zero product property for (x+13)(x-7)=0

umka21 [38]

(x+13)*(x-7)=0
Split into possible cases :
________________________
x+13=0
x-7=0
Solve the equations:
________________________
x= -13
x= 7
Final solution :

8 0

2 years ago

Read 2 more answers

A ship leaves port at noon and has a bearing of S29oW. The ship sails at 20 knots. How many nautical miles south and how many na

ira [324]

Answer:

Approximately $58.2\; \text{nautical miles}$ (assuming that the bearing is ${\rm S$29^{\circ}$W}$ .)

Step-by-step explanation:

Let $v$ denote the speed of the ship, and let $t$ denote the duration of the trip. The magnitude of the displacement of this ship would be $v\, t$ .

Refer to the diagram attached. The direction ${\rm S$29^{\circ}$W}$ means $29^{\circ}$ west of south. Thus, start with the south direction and turn towards west (clockwise) by $29^{\circ}$ to find the direction of the displacement of the ship.

The hypothenuse of the right triangle in this diagram represents the displacement of the ship, with a length of $v\, t$ . The dashed horizontal line segment represents the distance that the ship has travelled to the west (which this question is asking for.) The angle opposite to that line segment is exactly $29^{\circ}$ .

Since the hypotenuse is of length $v\, t$ , the dashed line segment opposite to the $\theta = 29^{\circ}$ vertex would have a length of:

$\begin{aligned}& \text{opposite (to $\theta$)} \\ =\; & \text{hypotenuse} \times \frac{\text{opposite (to $\theta$)}}{\text{hypotenuse}} \\ =\; & \text{hypotenuse} \times \sin (\theta) \\ =\; & v\, t \, \sin(\theta) \\ =\; & v\, t\, \sin(29^{\circ})\end{aligned}$ .

Substitute in $\begin{aligned} v &= 20\; \frac{\text{nautical mile}}{\text{hour}}\end{aligned}$ and $t = 6\; \text{hour}$ :

$\begin{aligned} & v\, t\, \sin(29^{\circ}) \\ =\; & 20\; \frac{\text{nautical mile}}{\text{hour}} \times 6\; \text{hour} \times \sin(29^{\circ}) \\ \approx\; & 58.2\; \text{nautical mile}\end{aligned}$ .